Tiêu đề 7. P2C7. HSC PREP Papers 26_With Solve.pdf ✅

†f.Z Av‡jvKweÁvb  HSC Prep Papers 1 mßg Aa ̈vq †f.Z Av‡jvKweÁvb Physical Optics Topicwise CQ Trend Analysis UwcK 2016 2017 2018 2019 2021 2022 2023 †gvU Av‡jvi e ̈wZPvi: aviYv, Bqs Gi wØ wPo cixÿv 2 Ñ 1 4 9 21 16 53 Av‡jvi AceZ©b Ñ 1 Ñ 1 Ñ 6 Ñ 8 AceZ©b †MÖwUs Ñ Ñ Ñ 1 Ñ Ñ Ñ 1 Av‡jvi mgeZ©b Ñ Ñ Ñ Ñ Ñ 3 Ñ 3 * we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| DÏxcKwU jÿ ̈ Ki: [Xv. †ev. 23] 1.5m P †K›`axq Pig 8.835mm 1mm S1 S2 wØ-wPi cixÿYwU‡Z 5890 A o Av‡jvK iwk¥ e ̈envi Kiv n‡jv| (K) mym1⁄2Z Drm Kx? (L) Zi1⁄2 gy‡Li cÖK...wZ Drm n‡Z `~iZ¡ wbf©iÑ e ̈vL ̈v Ki| (M) cici `ywU D3⁄4¡j †Wvivi `~iZ¡ wbY©q Ki| mgvavb: cici `ywU D3⁄4¡j †Wvivi `~iZ¡, x = D a = 5890  10–10  1.5 10–3  x = 8.835  10–4 m  cici `ywU D3⁄4¡j †Wvivi `~iZ¡ 8.835  10–4 m| (Ans.) (N) P we›`ywU‡Z †Kvb ai‡bi e ̈wZPvi cvIqv hv‡e, MvwYwZK e ̈vL ̈v Ki| mgvavb: c_ cv_©K ̈, x = xna D = 8.835  10–3  10–3 1.5  x = 5.89  10–6 m  S = 2  x = 2 5890  10–10  5.89  10–6   = 20 ; hv  Gi †Rvo ̧wYZK| A_©vr P we›`ywU‡Z MVbg~jK e ̈wZPvi cvIqv hv‡e| (Ans.) 2| Bqs-Gi wØwPo cixÿ‡Y e ̈eüZ Av‡jvK Dr‡mi Zi1⁄2‣`N© ̈ 5896 A  , wPo؇qi ga ̈eZ©x `~iZ¡ 2 mm Ges wPo I c`©vi j¤^ `~iZ¡ 1 m| cieZ©x‡Z wPo؇qi `~iZ¡ A‡a©K Ges wPo I c`©vi `~iZ¡ wØ ̧Y Kiv n‡jv| [iv. †ev. 23] (K) Zi1⁄2 gyL Kv‡K e‡j? (L) e ̈vwZPvi I AceZ©b Av‡jvK NUbv `ywUi gv‡S †g.wjK cv_©K ̈ Kx? e ̈vL ̈v K‡iv| (M) 1g †ÿ‡Î `kg D3⁄4¡j †Wvivi †K›`axq D3⁄4¡j †Wviv n‡Z `~iZ¡ wbY©q K‡iv| mgvavb: `kg D3⁄4¡j †Wvivi `~iZ¡, x10 = nD a = 10  5896  10–10  1 2  10–3  x10 = 2.948  10–3 m  `kg D3⁄4¡j †Wvivi †K›`axq D3⁄4¡j †Wviv n‡Z `~iZ¡ 2.948  10–3 m| (Ans.) (N) wØZxq †ÿ‡Î †Wviv cÖ ̄ cwieZ©b nqÑ MvwYwZK e ̈vL ̈v `vI| mgvavb: cÖ_g †ÿ‡Î, x1 = D1 2a1 = 5896  10–10  1 2  2  10–3  x1 = 1.474  10–4 m wØZxq †ÿ‡Î, x2 = D2 2a2 = 5896  10–10  2  1 2  2 2  10–3  x2 = 5.896  10–4 m  x1  x2 A_©vr wØZxq †ÿ‡Î †Wviv cÖ ̄’ cwieZ©b nq| (Ans.) 3| 1 mm S1 S2 1 m O †K›`axq D3⁄4¡j †Wviv M < N [Kz. †ev. 23] evqy gva ̈‡g Bqs-Gi wØwPo cixÿvq e ̈eüZ Av‡jvi Zi1⁄2 •`N© ̈  = 3800 A o Ges S2P – S1P = 6|
2  HSC Physics 2nd Paper Chapter-7 (K) †MÖwUs aaæeK Kv‡K e‡j? (L) d«bndvi †kÖwY AceZ©‡b DËj †jÝ e ̈envi Kiv nq †Kb? e ̈vL ̈v K‡i| (M) wP‡Î O Ges P we›`yi ga ̈Kvi `~iZ¡ wbY©q K‡iv| mgvavb: c_ cv_©K ̈ = n  S2P – S1P = n  n = 6  n = 6  x6 = 6D a = 6  3800  10–10  1 10–3  x6 = 2.28  10–3 m  O Ges P we›`yi ga ̈Kvi `~iZ¡ 2.28  10–3 m| (Ans.) (N) mgMÖ cixÿvwU 1.30 cÖwZmiv‡1⁄4i †Kv‡bv gva ̈‡g m¤úbœ Kiv n‡j 12Zg AÜKvi †Wvivi †K.wYK Ae ̄ v‡bi Kx cwieZ©b n‡e? MvwYwZKfv‡e we‡kølY K‡iv| mgvavb: cÖ_g †ÿ‡Î, AÜKvi †Wvivi Rb ̈, asin = (2n – 1)  2   = sin–1           2n – 1 2   a = sin–1           2  12 – 1 2  3800  10–10 10–3   = 0.25 wØZxq †ÿ‡Î,  =     = 3800  10–10 1.3 = 2.923  10–7 m  a sin = (2n – 1)  2   = sin–1           2n – 1 2   a = sin–1           2  12 – 1 2  2.923  10–7 10–3   = 0.193 ∵  <  A_©vr 1.3 cÖwZmiv‡1⁄4 12 Zg AÜKvi †Wvivi †K.wYK Ae ̄’vb c~‡e©i †P‡q n«vm cv‡e| (Ans.) 4| wP‡Î Bqs Gi wØ-wPo cixÿvi GKwU e ̈e ̄’v †`Lv‡bv n‡jv| wP‡o 3100 A o Zi1⁄2‣`‡N© ̈i Av‡jv †djv n‡j c`©vi †K›`a n‡Z Dfq w`‡K 10 wU †Wviv †`Lv †Mj| [h. †ev. 23] a = 0.4 mm S1 S2 P O x10 D = 1m (K) mym1⁄2Z Drm Kx? (L) †iW‡bi Aa©vqy 3.82 w`b ej‡Z Kx eySvq? (M) c`©vq 10 Zg D3⁄4¡j †Wvivi †K.wYK miY KZ? mgvavb: 10Zg D3⁄4¡j †Wvivi `~iZ¡, x10 = 10 D a = 10  3100  10–10  1 0.4  10–3  x10 = 7.75  10–3 m  †K.wYK miY,  = tan–1     x10 D = tan–1     7.75  10–3 1   = 0.444 A_©vr c`©vq 10Zg D3⁄4¡j †Wvivi †K.wYK miY 0.444| (Ans.) (N) DÏxc‡K wPo `ywUi e ̈eavb A‡a©K Kiv n‡j c`©vq †Wvivi msL ̈vi Kx cwieZ©b n‡e? MvwYwZK we‡kølY Ki| mgvavb: c~‡e© †Wviv msL ̈v, N = 10 + 1 + 10 = 21 cieZ©x‡Z, x10 = nD a  n = 7.75  10–3  0.4 2  10–3 3100  10–10  1  n = 5 †gvU †Wvivi msL ̈v, N = 5 + 5 + 1 = 11 ∵ N < N A_©vr wPo `ywUi e ̈eavb A‡a©K Kiv n‡j c`©vq †Wvivi msL ̈v c~‡e©i †P‡q n«vm cv‡e| (Ans.) 5| Bqs Gi wØwPo cixÿvq wPo؇qi ga ̈eZ©x e ̈eavb 1 mm Gi †_‡K c`©vi `~iZ¡ 1 m| e ̈eüZ Av‡jvi Zi1⁄2 •`N© ̈ 6000 A o | [P. †ev. 23] (K) Zi1⁄2 gyL Kx? (L) my-msMZ Dr‡mi •ewkó ̈ e ̈vL ̈v Ki| (M) DÏxc‡Ki c`©vq m„ó †Wviv ̧‡jvi cÖ ̄ KZ? mgvavb: †Wvivi cÖ ̄’, x = D 2a = 6000  10–10  1 2  10–3  x = 3  10–4 m  c`©vq m„ó †Wviv ̧‡jvi cÖ ̄’ 3  10–4 m| (Ans.) (N) DÏxc‡Ki wPo `ywUi ga ̈eZ©x `~iZ¡, e ̈eüZ Av‡jvi Zi1⁄2 •`‡N© ̈i wØ ̧Y n‡j c`©vq m‡e©v”P KqwU D3⁄4¡j †Wviv cvIqv m¤¢e? †Zvgvi DËi MvwYwZK we‡køl‡Y `vI| mgvavb: m‡e©v”P msL ̈K D3⁄4¡j †Wvivi Rb ̈  = 90  a sin = n  2 sin90 = n  n = 2  D3⁄4¡j †Wvivi msL ̈v = (2n + 1) = (2  2 + 1) = 5 wU A_©vr DÏxc‡Ki wPo `ywUi ga© ̈eZ©x `~iZ¡, e ̈eüZ Av‡jvi Zi1⁄2 •`‡N© ̈i wØ ̧Y n‡j c`©vq m‡e©v”P 5wU D3⁄4¡j †Wviv cvIqv m¤¢e| (Ans.)
†f.Z Av‡jvKweÁvb  HSC Prep Papers 3 6| S1 D = 1m S2 P O [P we›`y‡Z `kv cv_©K ̈ = 6, Av‡jvi K¤úv1⁄4 = 1016 Hz, S1S2 = 1 mm, cvwbi cÖwZmiv1⁄4 = 1.33|] wP‡Î Bqs Gi wØ-wPo cixÿvi Z_ ̈ †`qv n‡jv| [e. †ev. 23] (K) Av‡jvi mgeZ©b Kx? (L) Bqs-Gi wØwPo cixÿvq e ̈wZPvi Svj‡ii †K›`axq cwÆi J3⁄4¡j ̈ Ab ̈vb ̈ D3⁄4¡j †Wvivi †P‡q †ewk †Kb? (M) O I P we›`yi ga ̈Kvi `~iZ¡ wbY©q Ki| mgvavb: `kv cv_©K ̈ = 2   x  x =  2  6  x = 3  n = 3 A_©vr, P we›`y‡Z 3q D3⁄4¡j †Wviv Drcbœ n‡e|  †K›`axq D3⁄4¡j †Wviv n‡Z 3q D3⁄4¡j †Wvivi `~iZ¡, x3 = 3D a = 3  3  108 1016  1 10–3  x3 = 9  10–5 m A_©vr O I P we›`yi ga ̈Kvi `~iZ¡ 9  10–5 m| (Ans.) (N) cixÿvwU cvwb‡Z m¤úbœ Ki‡j †Wvivi cÖ‡ ̄ i cwieZ©b MvwYwZKfv‡e we‡kølY Ki| mgvavb: cÖ_g †ÿ‡Î, x1 = D 2a[] = 3  10–8  1 2  10–3 m  x1 = 1.5  10–5 m  = c f = 3  108 1016 m   = 3  10–8 m wØZxq †ÿ‡Î, x2 = D 2a = 2.256  10–8  1 2  10–3 m  x2 = 1.128  10–5 m w =  w = 3  10–8 1.33 m  w = 2.256  10–8 m ∵ x2 < x1 A_©vr cixÿvwU cvwb‡Z m¤úbœ Ki‡j †Wvivi cÖ ̄’ c~‡e©i Zzjbvq n«vm cv‡e| (Ans.) 7| c`v_©weÁvb j ̈v‡ei Bqs Gi wØwPo cixÿvq GKeY©x 5890 A o Zi1⁄2‣`‡N© ̈i Av‡jvK Drm wPoØq 0.8 mm e ̈eav‡b Ges c`©v wPoØq n‡Z 1 m `~i‡Z¡ Av‡Q| wigv c`©v‡K wPo؇qi w`‡K 5.2 cm mwi‡q Ges mxgv c`©v‡K wecixZ w`‡K 5.2 cm mwi‡q e ̈wZPvi m3⁄4v ch©‡eÿY K‡i| wigv †Wvivi cÖ‡ ̄’i cwieZ©b 0.02 mm †`Lj| [wm. †ev. 23] (K) Av‡jvi mgeZ©b Kv‡K e‡j? (L) d«bdvi †kÖwYi AceZ©‡b DËj †jÝ e ̈envi Kiv nq †Kb? e ̈vL ̈v K‡iv| (M) c`©vi cÖv_wgK Ae ̄ v‡b cÖwZwU †Wvivi cÖ ̄ wbY©q K‡iv| mgvavb: †Wvivi cÖ ̄’, x = D 2a = 5890  10–10  1 2  0.8  10–3  x = 3.681  10–4 m  c`©vi cÖv_wgK Ae ̄’v‡b cÖwZwU †Wvivi cÖ ̄’ 3.681  10–4 m| (Ans.) (N) wigvi cixÿvi †K›`axq D3⁄4¡j cwÆ n‡Z Z...Zxq AÜKvi cwÆi `~iZ¡ cÖv_wgK Ae ̄ vb †_‡K hZUzKz K‡g mxgvi cixÿvq ZZUzKz e„w× cvqÑ MvwYwZKfv‡e hvPvBc~e©K we‡kølY K‡iv| mgvavb: AÜKvi †Wvivi `~iZ¡, xn = (2n – 1) D 2a  x3 = (2  3 – 1) 5890  10–10  1 2  0.8  10–3 = 1.841  10–3 m wigvi cixÿvq, x 3 = (2n – 1) D 2a = (2  3 – 1) 5890  10–10  (1 – 0.052) 2  0.8  10–3  x 3 = 1.745  10–3 m  x = x3 – x 3 = (1.841  10–3 – 1.745  10–3 )  x = 9.6  10–5 m mxgvi cixÿvq, x 3 = (2n – 1) D 2a = (2  3 – 1) 5890  10–10 (1 + 0.52) 2  0.8  10–3  x 3 = 1.936  10–3 m  x = x 3 – x3  x = (1.936  10–3 – 1.841  10–3 )  x = 9.6  10–5 m ∵ x = x A_©vr wigvi cixÿvq †K›`axq D3⁄4¡j cwÆ n‡Z Z...Zxq AÜKvi cwUi `~iZ¡ cÖv_wgK Ae ̄’vb †_‡K hZUzKz Kg nq mxgvi cixÿvq ZZUzKz e„w× cvq| (Ans.) 8| Bqs Gi wØ-wPo cixÿvq 5890 A  Av‡jv e ̈env‡i wPiØq †_‡K 1.5 m `~‡i ̄’vwcZ c`©vq e ̈wZPvi m3⁄4v m„wó Kiv n‡jv| wPi؇qi ga ̈eZ©x `~iZ¡ 1 mm| [w`. †ev. 23; g. †ev. 23] (K) nvB‡Mb‡mi bxwZ †jL| (L) Aby‣`N© ̈ Zi1⁄2 mgeZ©xZ nq bv †Kb? e ̈vL ̈v Ki|
4  HSC Physics 2nd Paper Chapter-7 (M) cÖ_g D3⁄4¡j †Wvivi †K.wYK we ̄Ívi wbY©q Ki| mgvavb: cÖ_g D3⁄4¡j †Wvivi `~iZ¡, x1 = D a = 5890  10–10  1.5 10–3  x1 = 8.835  10–4 m   = tan–1     x1 D = tan–1     8.835  10–4 1.5   = 0.034  cÖ_g D3⁄4¡j †Wvivi †K.wYK we ̄Ívi = 2 = 2  0.034 = 0.067 A_©vr cÖ_g D3⁄4¡j †Wvivi †K.wYK we ̄Ívi 0.067| (Ans.) (N) wPi I c`©vi Ae ̄ vb AcwiewZ©Z †i‡L `kg D3⁄4¡j †Wvivi Ae ̄ v‡b 15Zg AÜKvi †Wviv m„wó Kiv hv‡e Kx? MvwYwZKfv‡e e ̈vL ̈v `vI| mgvavb: `kg D3⁄4¡j †Wvivi `~iZ¡, x10 = 10D a = 10  5890  10–10  1.5 10–3  x10 = 8.835  10–3 m †K›`axq D3⁄4¡j †Wviv n‡Z n Zg AÜKvi †Wvivi `~iZ¡, xn = (2n – 1) D a  x15 = (2n – 1) D a   = 2  10–3  8.835  10–3 (2  15 – 1)  1.5   = 4.062  10–10 m A_©vr 4.062  10–10 m Zi1⁄2‣`‡N© ̈i Av‡jvi e ̈envi K‡i `kg D3⁄4¡j †Wvivi Ae ̄’v‡b 15 Zg AÜKvi †Wviv m„wó Kiv hv‡e| (Ans.) 9| [Xv. †ev. 22; iv. †ev. 22] 0.2 mm S1 S2 1m c`©v Bqs Gi wØ-wPo cixÿvq 5800 Å Zi1⁄2‣`‡N© ̈i Av‡jv e ̈envi Kiv n‡q‡Q| cieZ©x‡Z wØ-wPowU 0.2 mm cÖ‡ ̄’i GKK wPo Øviv cÖwZ ̄’vcb Kiv n‡jv| (K) mZ ̈K mviwY Kx? [Aa ̈vq-10] (L) m~h© †_‡K AvMZ Av‡jvi Zi1⁄2gy‡Li cÖK...wZ Kxiƒc n‡e? e ̈vL ̈v Ki| (M) e ̈wZPvi Svj‡ii cÖ ̄ wbY©q Ki| DËi: Svj‡ii cÖ ̄’ = D a = 5800 × 10–10 × 1 0.2 × 10–3 = 2.9 mm (N) Dfq‡ÿ‡Î GKB Zi1⁄2‣`‡N© ̈i Av‡jvi Rb ̈ 1g Pi‡gi †K.wYK we ̄Ívi Awfbœ n‡e wK bv? MvwYwZKfv‡e e ̈vL ̈v Ki| mgvavb: †K›`axq Pig n‡Z n-Zg Pi‡gi †K.wYK e ̈eavb n n‡j, tann = xn D x1 = D a = 5800  10–10  1 0.2  10–3 m  x1 = 2.9  10–3 m tan1 = x1 D  1 = tan–1     2.9  10–3 1 = 0.166  †K.wYK we ̄Ívi = 21 = 2  0.166 = 0.332 AceZ©‡bi †ÿ‡Î, asinn = (2n + 1)  2  a sin1 = (2  1 – 1)  2  sin1 = 3  5800  10–10 2  0.2  10–3  1 = 0.249  †K.wYK we ̄Ívi, 21 = 2  0.249 = 0.498 A_©vr Dfq‡ÿ‡Î GKB Zi1⁄2‣`‡N© ̈i Av‡jvi Rb ̈ 1g Pi‡gi †K.wYK we ̄Ívi Awfbœ n‡e bv| (Ans.) 10| cixÿvMv‡i GKwU d«bndvi †kÖwYi GKK wP‡oi `iæb AceZ©b cixÿvq 5890Å Zi1⁄2‣`‡N© ̈i Av‡jv e ̈envi Kiv n‡jv| wØZxq μ‡gi Pi‡gi Rb ̈ AceZ©b †KvY 10 cvIqv †Mj Ges `kg Aeg we›`ywU cvIqvi †Póv Kiv n‡jv| [Kz. †ev. 22] (K) GK Kzj¤^ Kx? [Aa ̈vq-2] (L) `ywU we›`yi wefe cv_©K ̈ 6V ej‡Z Kx eySvq? [Aa ̈vq-3] (M) wP‡oi †ea wbY©q Ki| mgvavb: Pig we›`yi Rb ̈, asin = (2n + 1)  2  a = (2  2 + 1)  5890  10–10 2  sin10  a = 8.48  10–8 m  wP‡oi †ea 8.48  10–6 m| (Ans.) (N) cixÿ‡Y `kg Aeg we›`ywU cvIqvi †Póv mdj n‡qwQj wK? MvwYwZK hyw3mn †Zvgvi gZvgZ `vI| mgvavb: Aeg we›`yi Rb ̈, asin = m   = sin–1    m   a = sin–1     11  5890  10–10 8.48  10–6   = 49.823 > 90 A_©vr `kg Aeg we›`ywU cvIqvi †Póv mdj n‡qwQj| (Ans.)